Question

Find the net work done by friction on a box that moves in a complete circle of radius 0.40m on a uniform horizontal floor. The coefficient of kinetic friction between the floor and the box is 0.13, and the box weights 35.0N

Answer 1

Answer:

42/ 5 pi J

Explanation:

The word done by friction is given by

$F=\mu Nx$

where μ = coefficent of frction, N = weight of the box, and x = distance traveled.

Now in our case

μ = 0.13

N = 35.0 N

x = 0.40 * 2 pi (since it is the circumeference of the circle

Therefore,

$F=(0.13)(35.0)(0.40\cdot2\pi)$

which simplifies to give

$F=\frac{42}{5}\pi$

which is our answer!